Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $q = \dfrac{p + 7}{p^2 + 16p + 63} \times \dfrac{3p^2 - 21p - 90}{p - 10} $
First factor out any common factors. $q = \dfrac{p + 7}{p^2 + 16p + 63} \times \dfrac{3(p^2 - 7p - 30)}{p - 10} $ Then factor the quadratic expressions. $q = \dfrac {p + 7} {(p + 7)(p + 9)} \times \dfrac {3(p - 10)(p + 3)} {p - 10} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(p + 7) \times 3(p - 10)(p + 3) } { (p + 7)(p + 9) \times (p - 10)} $ $q = \dfrac {3(p - 10)(p + 3)(p + 7)} {(p + 7)(p + 9)(p - 10)} $ Notice that $(p + 7)$ and $(p - 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {3(p - 10)(p + 3)\cancel{(p + 7)}} {\cancel{(p + 7)}(p + 9)(p - 10)} $ We are dividing by $p + 7$ , so $p + 7 \neq 0$ Therefore, $p \neq -7$ $q = \dfrac {3\cancel{(p - 10)}(p + 3)\cancel{(p + 7)}} {\cancel{(p + 7)}(p + 9)\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $q = \dfrac {3(p + 3)} {p + 9} $ $ q = \dfrac{3(p + 3)}{p + 9}; p \neq -7; p \neq 10 $